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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

Just drafted this: Second Derivative

Comments/suggestions welcome!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,723

Hi MIF;

Works good and Happy New year!

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,259

Happy New Year MathsIsFun,

That page works nicely for me.

Do you have plans to extend it to show what f '' can be used for?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

bob bundy wrote:

Do you have plans to extend it to show what f '' can be used for?

I do NOW ...

... can you help?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,176

Hi MathsIsFun,

Nice Page! Works well with me!!!

Character is who you are when no one is looking.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

It would be nice to have a link to a calculator that can calculate these derivatives for the common functions and graph f, f' and f'' in different colors on the same set of axes.

I think such programs exist, but are probably part of a larger program or are not free. A simple

free link to a program that just does the derivatives and graphing would be nice.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

You mean this?

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

That's nice and a good site to remember. It only lacks a graph of f, f' and f'' on the same set of

axes.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,723

Hi noelevans;

Plot[{x^3,differentiate(x^3,x),differentiate(differentiate(x^3,x),x)},{x,-5,5}]

or

saving some typing

Plot[{x^3,D(x^3,x),D(D(x^3,x),x)},{x,-5,5}]

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Thanks bobbym,

I used copy and paste to put Plot[{x^3,D(x^3,x),D(D(x^3,x),x)},{x,-5,5}] into the input instead of

the differentiate command and it plotted all three. So that site does much more than just differentiate. Neat!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,723

Hi noelevans;

Your welcome!

Hi MIF;

Is it possible to get the equation for the tangent line of the crosshairs displayed?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

bobbym wrote:

Is it possible to get the equation for the tangent line of the crosshairs displayed?

As in "y=3.105x+2.09" ?

Or perhaps just the slope.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,723

Hi;

It is great that I have a choice! The equation would be really nice.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,259

bob bundy wrote:

Do you have plans to extend it to show what f '' can be used for?

I do NOW ...

... can you help?

Sure will.

Give me a few days.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,259

hi MathsIsFun

Here's a first installment.

Example.

Find the turning points, and identify if each is a maximum, minimum or point of inflection.

The turning points of a function are places on the curve where the gradient is momentarily zero.

First differentiate the function.

Solving this quadratic gives

Substituting these into the original function means the turning points are at (1/3 , -0.18519) and at (-1 , 1)

The first graphs below show the function in blue and the derivative function in red.

Notice that the two places where the red curve crosses the x axis line up with the turning points of the function.

Now, if you have the graph in front of you, it is obvious that the first is a maximum and the second a minimum.

But what if you have to prove it without referring to the graph ?

Differentiate the gradient function.

So the gradient of the red curve at x = -1 is negative. Without having to see the graph I know the red curve crosses the x axis from positive values to negative values as x increases from less than -1 to more than -1.

So I know the blue curve goes from a positive gradient, through zero, to a negative gradient. So it must be a maximum.

So the gradient of the red curve at x = 1/3 is positive. Without having to see the graph I know the red curve crosses the x axis from negative values to positive values as x increases from less than 1/3 to more than 1/3.

So I know the blue curve goes from a negative gradient, through zero, to a positive gradient. So it must be a minimum.

The red curve is repeated along with its gradient function (the double differentiated function) in blue on the second graphs below. I have put little + and - signs close to the turning point values to show how the red curve goes from + to - or from - to +

Rule for identifying maximums and minimums:

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

Great example, will adapt it to web page.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,259

Thank you.

I'm a bit dis-satisfied with the explanation of gradients of gradient graphs. It starts getting complicated at that stage. I've just had a thought. Maybe a question and answer stage before the final rule. I'll work on it tomorrow.

Would you like a piece on "points of inflexion" ?

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,259

hi MathsIsFun

Here is an alternative way of presenting the analysis of gradients as a series of questions. By making the reader answer step by step questions I think it leads them through the logic more carefully.

You could also add a close up picture of the relevant graph area.

I wrote:

Differentiate the gradient function.

So the gradient of the red curve at x = -1 is NEGATIVE.

Try to answer these questions:

Q1. Just to the left of x = -1 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

<include button with correct answer hidden = POSITIVE.>

Q2. Just to the right of x = -1 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

<include button with correct answer hidden = NEGATIVE.>

Now think about the original function.

Q3. Just to the left of x = -1 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

<include button with correct answer hidden = UPWARDS>

Q4. Just to the right of x = -1 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

<include button with correct answer hidden = DOWNWARDS>

Q5. So the turning point at x = -1 is a MAXIMUM / MINIMUM.

<inlcude button with correct answer hidden =MAXIMUM>

So the gradient of the red curve at x = 1/3 is POSITIVE.

Try to answer these questions:

Q6. Just to the left of x = 1/3 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

<include button with correct answer hidden = NEGATIVE.>

Q7. Just to the right of x = 1/3 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

<include button with correct answer hidden = POSITIVE.>

Now think about the original function.

Q8. Just to the left of x = 1/3 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

<include button with correct answer hidden = DOWNWARDS>

Q9. Just to the right of x = 1/3 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

<include button with correct answer hidden = UPWARDS>

Q10. So the turning point at x = 1/3 is a MAXIMUM / MINIMUM.

<inlcude button with correct answer hidden =MINIMUM>

i wrote:

Rule for identifying maximums and minimums: .................................

finish as before

Bob

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

Hi Bob,

I will have a look at that and get back to you (when I have completed a few other things I am doing now)

Yours, Rod

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**mathgogocart****Member**- Registered: 2012-04-29
- Posts: 1,426

Great,but make it a it longer.great calc.

Hey.

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